Consider the integral :$\int_0^1 {(tx+1-x)^n}dx$,n $\epsilon $ N and t is a parameter. Then, $$\int_0^1 {(tx+1-x)^n}dx=\frac{{t}^{n+1}-1}{{(n+1)}{(t-1)}}=\frac{[t^{n}+t^{n-1}+....+t+1]}{n+1}-----1$$ But,$${(tx+(1-x))^n}=\binom{n}{0}(tx)^n+\binom{n}{1}(tx)^{n-1}(1-x)+\binom{n}{2}(tx)^{n-2}(1-x)^2+...+\binom{n}{k}(tx)^{n-k}(1-x)^k+...+\binom{n}{n}(1-x)^n$$ So,$$\int_0^1 {(tx+1-x)^n}dx=\int_0^1\sum_{k=0}^{n}\binom{n}{k}(tx)^{n-k}.(1-x)^kdx=\sum_{k=0}^{n}t^{n-k}\int_0^1\binom{n}{k}.x^{n-k}(1-x)^kdx$$ using 1 :$$\frac{[t^{n}+t^{n-1}+....+t+1]}{n+1}=\sum_{k=0}^{n}t^{n-k}\int_0^1\binom{n}{k}.x^{n-k}(1-x)^kdx$$ Equating coefficients of $t^{n-k}$ on both sides: $$\frac{1}{n+1}=\binom{n}{k}\int_0^1x^{n-k}(1-x)^kdx=\binom{n}{k}\int_0^1(1-x)^{n-k}x^kdx$$ Hence,$$\int_0^1x^{k}(1-x)^{n-k}dx=\frac{1}{(n+1){\binom{n}{k}}}$$By taking $k=\alpha-1,n-k=\beta-1:$ $$B(\alpha,\beta)=\int_0^1x^{\alpha-1}(1-x)^{\beta-1}dx=\frac{{\Gamma(\alpha)}{\Gamma(\beta)}}{\Gamma(\alpha+\beta)}$$
Can anyone suggest me some other elementary methods for this that doesn't involves convolutions or double integrals ??